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For the binomial random graph $G(n,1/2)$, the chromatic number is $(1+o(1))n/(2{\log }_{2}n)$ with high probability. A first moment method bounds the independence number above. Edge-disjoint clique packing in the complement graph, followed by an edge-exposure martingale and a union bound over moderately large vertex sets, supplies independent sets for greedy colouring by removing independent sets.

If $X_t$ counts copies of the complete graph $K_t$ in $G(n,p)$, thenConsequently $p{n}^{2/(t-1)}\to 0$ implies $\mathbb{P}(X_t>0)\to 0$ by the first moment method.

Throughout this question the binomial random graph has $p=1/2$, as fixed in the introduction. Let $L={\log }_{2}n$ and $s=\lceil 2L\rceil$. Count the $s$-vertex independent sets by a random variable $I_s$. Linearity of expectation givesFor the second inequality, $s\ge 2L$ implies $sL-s(s-1)/2\le s/2\le L+1/2$. For the limit, $s!\ge (s/2{)}^{\lfloor s/2\rfloor }$ grows faster than $n$. By the first moment method, with high probability there is no such independent set; any larger independent set would contain one. Thus the independence number satisfies $\alpha (G)\le s-1<2L$ with high probability.

Every class of a proper graph coloring is an independent set, giving the claimed chromatic lower bound:

The Markov inequality states that for a nonnegative random variable $Z$ and $a>0$,Because $T$ is integer-valued, apply it with $a=1$. With $p=n^{-\alpha }$,ThereforeThis is the first moment method: if the expected triangle count in a binomial random graph tends to zero, triangles occur with vanishing probability.

Let $X_s$ count the copies of the complete graph $K_s$ in the binomial random graph $G_{n,p}$. By linearity of the expected value and independence of distinct edges,Throughout, $n$ is sufficiently large that $k=\lfloor \log \log n\rfloor \ge 2$. We have $k\to \infty$, $k=O(\log \log n)$ and $k\log k=o(\log n)$.

First, since $p=n^{-2/k}$,The first moment method proves $\mathbb{P}(X_{k+1}>0)=o(1)$.

For the existence of a $k$-clique, put $X=X_k$ and $\mu =\mathbb{E}X$. Since $p^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}=n^{-(k-1)}$,This alone is insufficient: we must control the variance via the overlap formula for the variance of a clique count. Let $I_S$ indicate that a given $k$-set $S$ spans a clique. If $|S\cap T|=j$, the two copies share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges, soFor $j=0,1$ their edge sets are disjoint and the covariance vanishes, even when they share one vertex. Counting ordered pairs of $k$-sets gives the exact formulaThis includes the diagonal case $j=k$.

To bound the overlap factor, regard it as the probability that a uniformly chosen $k$-set $T$ intersects a fixed $S$ in exactly $j$ vertices. A union bound over the possible specified $j$-subsets of $S$ yieldsHere $(a{)}_j=a(a-1)\cdots (a-j+1)$ is a falling factorial, and the last bound holds for $n\ge 2k$. ThereforeFor $2\le j\le k$, the concave quadratic $j(k-j+1)$ is minimized at an endpoint. Its endpoint values are $2(k-1)$ and $k$, both at least $k$. Thus every power of $n$ in the last sum is at most $n^{-1}$, andbecause the logarithm of its numerator is $O(k\log k)=o(\log n)$. The second moment method, or Chebyshev's inequality directly, gives $\mathbb{P}(X=0)=o(1)$. Finally, every larger clique contains a $(k+1)$-clique, so the absence of $K_{k+1}$ excludes all larger sizes. The clique number is exactly $k$ with high probability:

A $k$-colouring assigns one of $k$ colours to every vertex so that adjacent vertices have different colours. The chromatic number $\chi (G)$ is the least such $k$. An independent set contains no adjacent pair, and the independence number $\alpha (G)$ is the maximum size of such a set.

For a concrete first example, start with the five-cycle $C_5$ and apply the Mycielski construction $r-2$ times. Each application raises the chromatic number by one and preserves triangle-freeness. The resulting graph has chromatic number $r+1>r$ and contains no triangle, hence no complete graph $K_r$ for $r\ge 3$.

We next prove the stronger high-girth assertion by the probabilistic method. For a large integer $n$, takeIf $X$ counts cycles of lengths $3,\dots ,g$, thenBy the first moment method, with probability tending to one $X<n/2$.

Let $m=\lfloor n/(2k)\rfloor$. The expected number of independent sets of size $m$ is at mostbecause $p{m}^2=\Theta (n^{1+1/(2g)})$ dominates $m\log (en/m)=O(n)$. Hence with positive probability $G$ has fewer than $n/2$ short cycles and $\alpha (G)<m$.

Delete one vertex from each cycle of length at most $g$. The remaining graph $H$ has no such cycle, has at least $n/2$ vertices, and still has $\alpha (H)<n/(2k)$. Since each colour class is independent,This proves that there are graphs of arbitrarily high girth and chromatic number.

For the final claim, begin instead withThe expected number $T$ of triangles isso with probability tending to one $T<n$. Put $m=\lceil n^{0.7}\rceil$. The expected number of independent $m$-sets is bounded bybecause the positive term is $O(n^{0.7}\log n)$ whereas the negative term has order $n^{0.71}$.

Thus some such $G$ has fewer than $n$ triangles and no independent set of size $m$. Delete one vertex from every triangle; more than $n$ vertices remain and the graph is triangle-free. Take any induced subgraph on exactly $n$ vertices. Vertex deletion cannot increase the independence number, so the resulting graph satisfiesThis is the triangle-free graph with sub-power independence number construction.

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

Fix $k\ge 1$ and sample $G$ from the Erdős-Rényi model with $p=\frac{1}{2}{n}^{-2/3}$. Let $X$ count copies of the complete bipartite graph $K_{3,3}$. By the expected subgraph count in the Erdős-Rényi model,

Put $m=\lceil n/k\rceil$ and let $Z$ count independent set of size $m$. If the chromatic number satisfies $\chi (G)\le k$, some colour class has at least $m$ vertices, so $Z\ge 1$. Moreover,because the positive part of the exponent is $O(n)$ whereas $p\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ has order $n^{4/3}$. The first moment method therefore gives $\mathbb{P}(\chi (G)\le k)\le \mathbb{E}Z=o(1)$.

Using Markov inequality for $X$ and the union bound,for all sufficiently large $n$. Hence at least one such $G$ contains no $K_{3,3}$ and has $\chi (G)>k$.

Let $Y$ count unordered pairs of distinct $K_4$ copies that have a common vertex. Classifying such a pair by whether it shares one, two, or three vertices givesIndeed, after substituting $p=n^{-2/3}\log n$, these terms are respectively

The first moment method gives $\mathbb{P}(Y>0)\le \mathbb{E}Y\to 0$. Part (i) says that $X\ge 100$ with probability tending to one, so the union bound shows that, with probability tending to one, both $X\ge 100$ and $Y=0$. On that event any one hundred of the $K_4$ copies are pairwise vertex-disjoint, proving vertex-disjoint sparse clique copies in this case.

A connected graph on $s$ vertices containing at least two cycles has a spanning tree together with at least two further edges. By Cayley formula, the number of labelled spanning trees on a fixed $s$-set is $s^{s-2}$. Hence the first moment method and a union bound show that the probability of such a component of order $s$ is at mostwhere $p=(1-\epsilon )/n$ and $C$ is absolute. Requiring that the chosen set be a component would only add absent-edge conditions, so omitting them is a valid upper bound.

ThereforeThus every component in the stated range is either a tree or a unicyclic component with high probability.